= Solution
The set is the <kernel> of the <linear map> $L(\mathbf r)=\mathbf v\cdot\mathbf r$. It contains zero, and <linearity> gives $L(\alpha\mathbf r+\beta\mathbf s)=0$ whenever $L(\mathbf r)=L(\mathbf s)=0$, proving closure under <linear combinations>. If $\mathbf v\ne0$, it is a plane through the origin; if $\mathbf v=0$, it is all of $\mathbb R^3$. \b[An arbitrary plane need not be a vector subspace]: for example $z=1$ omits the zero vector. An <affine subspace> and a vector subspace are not interchangeable.
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